Question

For the following exercises, evaluate the integrals, if possible.\(\int_{1}^{\infty} \frac{e^{-x}}{x} d x\)

Short answer

The integral converges.

Step-by-step solution

Recognize the Type of Integral

The given integral is \( \int_{1}^{\infty} \frac{e^{-x}}{x} \, dx \), which is an improper integral because it has an infinite limit of integration (\( \infty \)).

Consider Convergence

To determine if the integral converges or diverges, it's important to analyze the behavior of the function as \( x \to \infty \). Since \( e^{-x} \to 0 \) and \( \frac{1}{x} \to 0 \), we should formally analyze this using a technique like the Comparison Test.

Apply the Comparison Test

Use a known convergent integral for comparison. Note that \( e^{-x} \leq 1 \) for all \( x \geq 1 \), so \( \frac{e^{-x}}{x} \leq \frac{1}{x} \). \( \int_{1}^{\infty} \frac{1}{x} \, dx \) diverges (harmonic). As \( e^{-x} \) decreases rapidly enough, conduct a limit comparison with the integral \( \int_{1}^{\infty} e^{-x} \, dx \), which converges.

Evaluate the Limit Comparison Integral

Compute \( \int_{1}^{\infty} e^{-x} \, dx \) directly. By substitution, this integral equals \([-e^{-x}]_1^{\infty} = 0 - (-e^{-1}) = e^{-1}\), which converges.

Conclude Convergence of the Improper Integral

Since \( \int_{1}^{\infty } e^{-x} \, dx \) converges and \( \frac{e^{-x}}{x} \) is smaller than \( e^{-x} \) for all \( x > 0 \), the original integral \( \int_{1}^{\infty} \frac{e^{-x}}{x} \, dx \) converges by comparison.

Key concepts

Convergence Tests

When dealing with improper integrals, it's crucial to determine whether they converge or diverge. Convergence tests help us understand this behavior. For an integral to converge, the area under the curve of the function from a certain point to infinity must be finite.
To decide on convergence, one popular method is the **Comparison Test**. This requires us to compare our function with another function that we already know converges or diverges. If the function of interest is smaller than a known convergent function for sufficiently large values of the variable, it also converges. Conversely, if it is larger than a known divergent function, it also diverges.
Another useful method is the **Limit Comparison Test**. This test involves taking the limit of the ratio of the two functions. If this limit is positive and finite, the two integrals will behave the same way—either both converge or both diverge. These methods make it easier to predict the behavior of complex integrals without direct computation.

Limit Comparison Test

The Limit Comparison Test is a powerful tool for evaluating improper integrals. It's especially handy when it isn't immediately clear how an integral behaves. This test offers a way to draw connections between the behavior of different functions.
Here’s a quick breakdown of how it works:

• Identify a function, say \( b(x) \), with known behavior relative to your integral. This can be either convergent or divergent.
• Calculate the limit \( \lim_{x \to \infty} \frac{a(x)}{b(x)} \), where \( a(x) \) is the function you are examining.
• If this limit results in a positive finite number, both integrals \( \int a(x) \, dx \) and \( \int b(x) \, dx \) will behave the same in terms of convergence.

For our original problem, comparing \( \frac{e^{-x}}{x} \) with \( e^{-x} \) helped us determine that since \( \int e^{-x} \, dx \) converges, so does \( \int \frac{e^{-x}}{x} \, dx \). This illustrates how the Limit Comparison Test can simplify the analysis of complex integral behavior.

Exponential Functions

Exponential functions play a key role in many mathematical contexts, including evaluating integrals. The function \( e^{-x} \) is a crucial example due to its unique properties.
Exponential decay, represented by \( e^{-x} \), implies that as \( x \) increases, \( e^{-x} \) approaches zero very quickly. This rapid decrease makes \( e^{-x} \) frequently useful in simplifying integrals, especially those involving infinity.
The integral of \( e^{-x} \) over an infinite interval

• implies convergence, since the area under the curve becomes finite.
• This specific property makes it suitable for use in several comparison tests when evaluating other related functions.

Understanding these characteristics of exponential functions helps in effectively applying them in convergence tests and other mathematical analyses.